When regrouping, you're essentially simplifying the problem into smaller parts. For instance, you would calculate 60 + 40=? and then solve 4 + 3=? The final answer would be 107.
Let's convert each statement into mathematical terms.
Pineapple = p
Orange = x
is indicates equality
Pineapple has a weight that is 7 times that of an orange: p = 7x
Pineapple is 870 grams heavier than the orange. p = 870 + x
Insert p = 7x into the second equation:
7x = 870 + x
Subtract x from both sides
6x = 870
Divide by 6
x = 145
The weight of the orange is 145 grams.
The weight of the pineapple is 7 times that of the orange = 7(145) = 1015 grams.
Assuming we start with a full standard deck of 52 and then draw 4 spades along with 1 card from a different suit, that leaves us with 47 cards still in the deck. We are hoping to draw another spade from these remaining cards. Initially, there were 13 spades, but after drawing five cards, 9 spades remain in the deck. The likelihood of pulling one of these 9 spades from the 47 cards is
In other words, we want to get any 1 of the available 9 spades while avoiding any of the other 38 non-spades, and we're drawing just a single card from the 47 cards total.
Response:
a. 55 cars
b. 25 cars
Detailed explanation:
Let’s denote the quantity of cars with stereo systems as N(ss), those with air conditioning as N(ac), and those with sunroofs as N(sr).
We find that:
N(ss) = 30
N(ac) = 30
N(sr) = 40
N(ss and ac and sr) = 15
N(at least two) = 30
a.
To calculate how many cars possess at least one feature (N(at least one) or N(ss or ac or sr)), we apply:
N(ss or ac or sr) = N(ss) + N(ac) + N(sr) - N(ss and ac) - N(ss and sr) - N(ac and sr) + N(ss and ac and sr)
N(ss or ac or sr) = 30 + 30 + 40 - (N(at least two) + 2*N(ss and ac and sr)) + 15
Substituting, we find N(ss or ac or sr) = 30 + 30 + 40 - (30 + 2*15) + 15 = 55
b.
For those cars that have exactly one feature, we have:
N(only one) = N(at least one) - N(at least two)
N(only one) = 55 - 30 = 25
